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### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2011

MCQ (Single Correct Answer)
A Carnot engine operating between temperatures ${{T_1}}$ and ${{T_2}}$ has efficiency ${1 \over 6}$. When ${T_2}$ is lowered by $62$ $K$ its efficiency increases to ${1 \over 3}$. Then ${T_1}$ and ${T_2}$ are, respectively:
A
$372$ $K$ and $330$ $K$
B
$330$ $K$ and $268$ $K$
C
$310$ $K$ and $248$ $K$
D
$372$ $K$ and $310$ $K$

## Explanation

Efficiency of engine
${1 \over 6} = 1 - {{{T_2}} \over {{T_1}}}$ and ${\eta _2} = 1 - {{{T_2} - 62} \over {{T_1}}} = {1 \over 3}$
$\therefore$ ${T_1} = 372\,K$ and ${T_2} = {5 \over 6} \times 372 = 310\,K$
2

### AIEEE 2011

MCQ (Single Correct Answer)
Three perfect gases at absolute temperatures ${T_1},\,{T_2}$ and ${T_3}$ are mixed. The masses of molecules are ${m_1},{m_2}$ and ${m_3}$ and the number of molecules are ${n_1},$ ${n_2}$ and ${n_3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:
A
${{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}$
B
${{{n_1}T_1^2 + {n_2}T_2^2 + {n_3}T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}$
C
${{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}$
D
${{\left( {{T_1} + {T_2} + {T_3}} \right)} \over 3}$

## Explanation

Number of moles of first gas $= {{{n_1}} \over {{N_A}}}$
Number of moles of second gas $= {{{n_2}} \over {{N_A}}}$
Number of moles of third gas $= {{{n_3}} \over {{N_A}}}$
If there is no loss of energy then
${P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV$
${{{n_1}} \over {{N_A}}}R{T_1} + {{{n_2}} \over {{N_A}}}R{T_2} + {{{n_3}} \over {{N_A}}}R{T_3}$
$= {{{n_1} + {n_2} + {n_3}} \over {{N_A}}}R{T_{mix}}$
$\Rightarrow {T_{mix}} = {{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}$
3

### AIEEE 2010

MCQ (Single Correct Answer)
A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32$ $V$, the efficiency of the engine is
A
$0.5$
B
$0.75$
C
$0.99$
D
$0.25$

## Explanation

${T_1}{V^{\gamma - 1}} = {T_2}{\left( {32V} \right)^{\gamma - 1}}$
$\Rightarrow {T_1} = {\left( {32} \right)^{\gamma - 1}}.{T_2}$
For diatomic gas, $\gamma = {7 \over 5}$
$\therefore$ $\gamma - 1 = {2 \over 5}$
$\therefore$ ${T_1} = {\left( {32} \right)^{{2 \over 5}}}.{T_2} \Rightarrow {T_1} = 4{T_2}$
Now, efficiency $= 1 - {{{T_2}} \over {{T_1}}}$
$= 1 - {{{T_2}} \over {4{T_2}}}$
$= 1 - {1 \over 4}$
$= {3 \over 4}$
$= 0.75.$
4

### AIEEE 2009

MCQ (Single Correct Answer)
Two moles of helium gas are taken over the cycle $ABCD$, as shown in the $P$-$T$ diagram.

The net work done on the gas in the cycle $ABCDA$ is:

A
$276$ $R$
B
$1076$ $R$
C
$1904$ $R$
D
zero

## Explanation

The net work in the cycle $ABCD$ is
$W = {W_{AB}} + {W_{BC}} + {W_{CD}} + {W_{DA}}$
$= 400R + 2.303nRT\log {{{P_B}} \over {{P_C}}} + \left( { - 400R} \right) - 414R$
$= 2.303 \times 2R \times 500\log {{2 \times {{10}^5}} \over {1 \times {{10}^5}}} - 414R$
$= 693.2\,R - 414\,R = 279.2\,R$

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